# Riemannian vs Non-Riemannian curvature

If you neither know the metric nor the holonomy group, how do you recognize a curvature tensor is Riemannian?

I assume a curvature, by definition, satisfies Bianchi identities. I know it is Riemannian if there exists a symmetric non degenerate tensor $g_{ab}$ such that these satisfy the condition $$g_{ea}R^e{}_{bcd}+g_{eb}R^e{}_{acd}=0 \, ,$$ but its solutions are not unique for the metric (a homogeneous equation).

In $n$ dimensions, these $\frac{n(n+1)}{2}\frac{n(n-1)}{2}$ conditions bring us down from the $\frac{n^2(n+1)(n-1)}{3}$ components of the curvature to $\frac{n^2(n+1)(n-1)}{3\cdot 2 \cdot 2}$ of the Riemann Tensor. Yet it seems to me these equations don't tell me what the metric should be.

• If the curvature vanishes, these equations don't say anything, so clearly they don't determine the metric. – Ben McKay Jan 4 '18 at 18:41
• Not all curvatures are Riemannian, if you don't know the metric how can you tell? – Aureliano Skirzewski Jan 4 '18 at 19:07
• @BenMcKay, you are right but, If the curvature is zero at a point, that would just tell me that any non-degenerated tensor can be parallel transported to be the metric elsewhere (if parallel transport of that tensor doesn't depend on the path). – Aureliano Skirzewski Jan 4 '18 at 19:22
• Is your question different from that answered at mathoverflow.net/questions/202211/… ? – macbeth Jan 4 '18 at 19:23
• It is not the same question, in that post @orbit assumes a tensor with Young tableu (2,2). It is interesting but it assumes wrongly the Riemann is a (0,4) instead of a (1,3) tensor. I rather ask if there is a metric that would turn my (1,3) tensor into a (0,4) with Young tableu (2,2) – Aureliano Skirzewski Jan 4 '18 at 20:07

## 1 Answer

NB: In what follows, to save typing, I will be working on a manifold $M$, but I will write $T$, $T^*$, etc. to denote the bundles $TM$, $T^*M$, etc. and let $M$ be understood.

It seems that the OP wants to be able to test whether a $(1,3)$-tensor $R$ with the symmetries of a Riemann curvature tensor is actually the curvature of a Riemannian metric. By the symmetries of the Riemann curvature tensor, we mean that $R$ is a section of $T\otimes T^*\otimes \Lambda^2T^*$ that is in the kernel $B$ of the natural skew-symmetrization mapping $$T\otimes T^*\otimes \Lambda^2T^* \longrightarrow T\otimes \Lambda^3T^*.$$

There are local and global aspects of this problem, even in dimension $2$, and there are degeneracy problems that show up when the tensor is allowed to vanish at some places.

For example, the OP's response to Ben's comment about the vanishing of the curvature at a point doesn't quite make sense because there is no connection specified in the problem, just the tensor $R$, so there is no parallel transport available to transport either $g$ or $R$ to other places.

Of course, the first algebraic condition is that there must be a nondegenerate section $g$ of $S^2T^*$, for which the natural 'index-lowering' pairing $\langle g, R\rangle$ (which, a priori takes values in $T^*\otimes T^*\otimes \Lambda^2 T^*$) takes values in $\Lambda^2T^*\otimes\Lambda^2T^*$. As the OP remarks, this condition does not determine a $g$ uniquely, since, for example, one can always replace such a $g$ by a multiple of $g$.

The existence of any such nondegenerate $g$ does put algebraic conditions on $R$, though, since, for example, it implies that $R$ must take values in $(T\otimes T^*)_0\otimes \Lambda^2 T^*$, where $(T\otimes T^*)_0\subset T\otimes T^*$ is the space of endomorphisms of trace $0$. Call the corresponding subbundle $B_0\subset B$. Of course, there are further conditions, in order that there be a nondegenerate $g$ for which $\langle g, R\rangle$ takes values in $\Lambda^2T^*\otimes\Lambda^2T^*$, even in dimension $n=2$.

For example, in dimension $2$, if $R$ is a nonzero section of $B_0$ (which has rank $3$) and $R$ happens to take values in the subset $N\otimes \Lambda^2T^*$, where $N\subset (T\otimes T^*)_0$ is the set of (nonzero) nilpotent endomorphisms, then there is no nondegenerate $g$ satisfying the given condition. On the other hand, if $R$ is a nonzero section of $B_0$ and nowhere takes values in $N\otimes \Lambda^2 T^*$, then, locally, there always exists a $g$ whose Riemann curvature tensor is $R$. In fact, the local solutions depend on $2$ functions of one variable. However, there still might not be any global solutions. For example, if $g_0$ is the standard metric on the $2$-sphere with Gauss curvature $+1$ and $R_0$ is its $(1,3)$-curvature tensor, then $-R_0$ cannot be the curvature tensor of any nondegenerate $g$ globally defined on $S^2$.

However, the real problems start when $n=3$. If $R$ is a section of $B_0$, then we can form its space of curvature endomorphisms $R(X,Y)$ where $X$ and $Y$ are vector fields. These, of course, take values in $(T\otimes T^*)_0$, and, at each point $p\in M$, they span a subspace $E(R_p)\subset (T_p\otimes T^*_p)_0$ of dimension at most $3$. Suppose that $R$ satisfies the generic condition that this span has dimension $3$ at every point $p$. Then, at each point $p$, the subspace $E(R_p)$ must span a simple Lie algebra of dimension $3$ in $(T\otimes T^*)_0$, either isomorphic to $\mathfrak{so}(3)$ or $\mathfrak{so}(2,1)$. This is a further pointwise algebraic condition on $R$. Moreover, in either of these cases, the space of 'compatible' nondegenerate quadratic forms $g$ that satisfy the condition that $\langle g, R\rangle$ take values in $\Lambda^2T^*\otimes\Lambda^2T^*$ are all multiples of a single nondegenerate quadratic form $g_0$ (of signature $(3,0)$ or $(2,1)$. Thus, the problem you are trying to solve is whether there is a nonzero function $u$ so that the nondegenerate quadratic form $g = u g_0$ has $R$ as its curvature. This is a wildly overdetermined problem, with at most a finite-dimensional space of solutions even locally. Thus, even with the algebraic necessary condition on $E(R)$, you cannot expect to solve this problem, even locally, although determining the existence of a solution is essentially reduced to an ODE problem that can be solved algorithmically. I can supply details if you are interested.

There are more subtle cases, when the rank of $E(R)$ at each point is $1$ or $2$, but, again, since the explicit analysis is long, let me save that unless there is interest.

For $n>3$, there are, again, many further algebraic conditions and then differential conditions involved in the test for when a $(1,3)$-tensor satisfying the symmetries of a Riemann curvature tensor is actually a Riemann curvature tensor, but the calculations rapidly become unmanageable, though they are theoretically doable.

• I assumed there is a connection when I said it satisfies Bianchi identities. I focused on solving the problem of the algebraic restrictions on the curvature but my problem is with a connection. I thought of rephrasing to a question of whether a connection is Levi-Civita, but thinking in those terms I found an older question mathoverflow.net/questions/54434/… which you answered too. I believe the answer I was looking for may be in that other post. Thanks! – Aureliano Skirzewski Jan 5 '18 at 13:38
• @AurelianoSkirzewski: No, your question is different from the one that I was answering in that other question because, as you stated, you only assumed that you were given $R$ as a $(1,3)$-tensor satisfying the first Bianchi identity (which is purely algebraic and does not involve a connection). Your question, as I understood it, was whether or not there was any metric $g$ for which $R$ was the Riemann curvature tensor. There is a separate, but weaker, question of whether or not $R$ is the curvature of some torsion-free connection, which it might be, even if there is no compatible metric $g$. – Robert Bryant Jan 5 '18 at 13:59